Two radioactive materials A and B have decay constants $$10\lambda$$ and $$\lambda$$, respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of A to that of B will be 1/e after a time:

For any radioactive material, the number of undecayed nuclei at a time $$t$$ is given by the exponential-decay law

$$N(t)=N_0\,e^{-\lambda t},$$

where $$N_0$$ is the initial number of nuclei and $$\lambda$$ is the decay constant of that material.

We are told that the two samples A and B start with the same initial number of nuclei, so we can write for each sample

$$N_A(0)=N_B(0)=N_0.$$

The decay constants are different: for A it is $$10\lambda$$ and for B it is $$\lambda$$. Using the decay formula separately for the two samples we have

$$N_A(t)=N_0\,e^{-(10\lambda)t}$$

and

$$N_B(t)=N_0\,e^{-(\lambda)t}.$$

We are interested in the ratio of the numbers of nuclei remaining after time $$t$$:

$$\frac{N_A(t)}{N_B(t)}=\frac{N_0\,e^{-(10\lambda)t}}{N_0\,e^{-(\lambda)t}}.$$

Because the same $$N_0$$ appears in numerator and denominator, it cancels out, leaving

$$\frac{N_A(t)}{N_B(t)}=e^{-(10\lambda)t}\,e^{+(\lambda)t}=e^{-(10\lambda t-\lambda t)}=e^{-9\lambda t}.$$

The problem states that this ratio becomes $$\dfrac{1}{e}$$, which is the same as $$e^{-1}$$. Therefore we set

$$e^{-9\lambda t}=e^{-1}.$$

Since the exponential function is one-to-one, the exponents themselves must be equal. Hence

$$-9\lambda t=-1.$$

Dividing both sides by $$-9\lambda$$, we obtain

$$t=\frac{1}{9\lambda}.$$

This value matches exactly the time given in Option B.

Hence, the correct answer is Option B.